What is the outer product of a tensor product of vectors?

  • #1
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Main Question or Discussion Point

If one has two single-particle Hilbert spaces ##\mathcal{H}_{1}## and ##\mathcal{H}_{2}##, such that their tensor product ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## yields a two-particle Hilbert space in which the state vectors are defined as $$\lvert\psi ,\phi\rangle =\lvert\psi\rangle\otimes\lvert\phi\rangle\in\mathcal{H}_{1}\otimes\mathcal{H}_{2}$$ where ##\lvert\psi\rangle\in\mathcal{H}_{1}## and ##\lvert\phi\rangle\in\mathcal{H}_{2}##.

Now, the inner product for ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## is defined such that $$\langle\phi ,\psi\vert\psi ,\phi\rangle =\left(\langle\phi\rvert\otimes\langle\psi\lvert\right)\left(\lvert\psi\rangle\otimes\lvert\phi\rangle\right) =\langle\psi\lvert\psi\rangle_{1}\langle\phi\lvert\phi\rangle_{2}$$ where ##\langle\cdot\lvert\cdot\rangle_{1}## is the inner product defined on ##\mathcal{H}_{1}## and ##\langle\cdot\lvert\cdot\rangle_{2}## the inner product defined on ##\mathcal{H}_{2}##.

How though is the outer product defined? Is it simply $$\lvert\psi ,\phi\rangle\langle\phi ,\psi\rvert =\lvert\psi\rangle\otimes\lvert\phi\rangle\langle\phi\rvert\otimes\langle\psi\rvert =\lvert\psi\rangle\langle\psi\rvert_{1} \lvert\phi\rangle\langle\phi\rvert_{2}$$ where ##\lvert\psi\rangle\langle\psi\rvert_{1}## is the outer product in ##\mathcal{H}_{1}## and ##\lvert\phi\rangle\langle\phi\rvert_{2}## is the outer product in ##\mathcal{H}_{2}##.
 

Answers and Replies

  • #2
andrewkirk
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It might be easier to think of more general outer products, rather than the outer product of an element with itself, as you have written.
That is, consider the outer product ##|a,b\rangle\langle c,d|=|a\rangle\otimes |b\rangle\langle c|\otimes\langle d|##.

This is an operator on the product space ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}##. To see what it does, we apply it to an element of that space ##|e,f\rangle=|e\rangle\otimes |f\rangle##.

The result is ##|a,b\rangle \langle c|e\rangle \langle d|f\rangle##
 
  • #3
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This is an operator on the product space H1⊗H2\mathcal{H}_{1}\otimes\mathcal{H}_{2}. To see what it does, we apply it to an element of that space |e,f⟩=|e⟩⊗|f⟩|e,f\rangle=|e\rangle\otimes |f\rangle.

The result is |a,b⟩⟨c|e⟩⟨d|f⟩
So one can't define it as ##\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\lvert b\rangle\langle d\rvert## then? Should it be something like $$\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\otimes\lvert b\rangle\langle d\rvert$$

I ask in particular as I'm trying to understand the notion of a partial trace.
 
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  • #4
andrewkirk
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So one can't define it as ##\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\lvert b\rangle\langle d\rvert## then?
That symbol string on the RHS is not intrinsically meaningful. It juxtaposes the two operators ##\lvert a\rangle\langle c\rvert## and ##\lvert b\rangle\langle d\rvert## but:
- since they are not scalars, the juxtaposition cannot be interpreted as scalar multiplication
- since they are operators on different spaces (##\mathcal H_1## and ##\mathcal H_2##), the juxtaposition cannot be interpreted as composition of operation.

Should it be something like $$\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\otimes\lvert b\rangle\langle d\rvert$$
That is one of a number of intrinsically meaningful ways of writing it. It has the same meaning as what I wrote above ##(
|a\rangle\otimes |b\rangle)(\langle c|\otimes\langle d|)## (I have added parentheses here that were only implied above, to make it clear what operations are being performed).
 
  • #5
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That is one of a number of intrinsically meaningful ways of writing it. It has the same meaning as what I wrote above (|a⟩⊗|b⟩)(⟨c|⊗⟨d|)
How does one show that the two expressions are equivalent? Would it be something like this:
$$\left(\lvert a\rangle\otimes\lvert b\rangle\langle c\rvert\otimes\langle d\rvert\right)\lvert e\rangle\otimes\lvert f\rangle =\lvert a\rangle\otimes\lvert b\rangle\left(\langle c\vert e\rangle\langle d\vert f\rangle\right)\\ =\left(\lvert a\rangle\langle c\vert e\rangle\right)\otimes \left(\lvert b\rangle\langle d\vert f\rangle\right) =\left(\lvert a\rangle\langle c\lvert\otimes\lvert b\rangle\langle d\lvert\right)\lvert e\rangle\otimes\lvert f\rangle$$
 
  • #6
andrewkirk
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I would write it as follows:

$$(|a,b\rangle\langle c,d|)|e,f\rangle\equiv (|a\rangle\otimes|b\rangle)(\langle c|\otimes\langle d|)(|e\rangle\otimes|f\rangle)=(\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)$$
and
$$(|a\rangle\langle c|\otimes|b\rangle\langle d|)|e,f\rangle\equiv
(|a\rangle\langle c|\otimes|b\rangle\langle d|)(|e\rangle\otimes |f\rangle)\
=(|a\rangle\langle c|)|e\rangle\otimes (|b\rangle\langle d|)|f\rangle
\equiv (\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)
$$
So
$$(|a,b\rangle\langle c,d|)=(|a\rangle\langle c|\otimes|b\rangle\langle d|)$$
 

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